Question

A 100.0 mL sample of a solution that is 0.125 M in HClO and 0.130 M in NaClO has added to it 2.00 mL of 0.100 M HCl. The Ka for HClO is 2.9 x 10-8. What is the pH of the resulting solution?

To start, I found the moles of each compound, then I added the moles of HClO and HCl to get my total moles of acid. Finally, I plugged it into the Henderson-Hasselbach equation.

I found that 0.0002 moles of HCl was added to a solution that contained 0.0125 moles of HClO and 0.013 moles of NaClO

So...

-log(2.9 -log(`2.9*10^-8`) + log(`0.013/0.0127`)
pH=7.55

Does this seem right? I wasn't sure how to approach this problem and I am unsure about the answer I got.

Thanks for any help!

Answer 1

`"pH" = 7.554`

Explanation:

Start by finding the quantity of each species in the `"NaClO"` buffer immediately after the addition of `"HCl"`.

• `n("HCl") = c * V = 2.00 xx 10^(-4) color(white)(l) mol`
• `n("HClO") = c * V = 1.25 xx 10^(-2) color(white)(l) mol`
• `n("NaClO") = c * V = 1.30 xx 10^(-2) color(white)(l) mol`

Hydrochloric acid `"HCl"` would react with `"NaClO"` to produce a weak electrolyte `"HClO"` through the following double decomposition reaction:

`stackrel((color(grey)("strong")))("HCl"(aq)) + "NaClO" (aq) color(navy)(to) stackrel((color(grey)("weak")))("HClO"(aq)) + "NaCl"(aq)`

`"NaClO"` is in excess and therefore

`n("HClO","final") = n("HClO","initial")+ n("HCl")`
`color(white)(n("HClO","final")) = 1.2502 xx 10^(-2) color(white)(l) mol`

`n("NaClO", "final") = n("NaClO", "initial") - n("HCl")`
`color(white)(n("NaClO","final")) = 1.2998 xx 10^(-2) color(white)(l) mol`

The rest of `"NaClO"` completely disassociates to produce `"Na"^(+)` and `"ClO"^(-)`.

`"NaClO"(s) color(navy)(to) "Na"^(+)(aq) + "ClO"^(-)(aq)`

`n("ClO"^(-)) = n("NaClO") = 1.30 xx 10^(-2) color(white)(l) mol`

The hydrolysis of aqueous `"ClO"^(-)` from `"NaClO"` might pose an influence on the acidity (in other words, `"pH"`) of the solution. A "RICE" table can help with the derivation of the equilibrium proton concentration due to this reversible process.

R `" ClO"^(-)(aq) + "H"_2"O"(l) color(darkgreen)(rightleftharpoons) "OH"^(-) (aq) + "HClO"(aq)`
I `color(purple)(color(white)(ll)1.2998 xx 10^(-2) color(white)(-------) color(white)(ll)1.2502 xx 10^(-2)`
C `color(darkgreen)(" "-x color(white)(-------) +x color(white)(---) +x)`
E `color(navy)(color(white)(ll)1.2998 xx 10^(-2) color(white)(-------)1.2502 xx 10^(-2))`
`color(navy)(color(white)(--) -x color(white)(-----lll-) +x color(white)(---)+x)`
(all values in this table are in `mol`)

`K_a = 2.9 xx 10^(-8)` for `"HClO"` such that
`K_b = (1.0 xx 10^(-14))/ (2.9 xx 10^(-8)) = 3.448 xx 10^(-7)` for `"ClO"^(-)`, its conjugate base.

#(n("OH"^(-)) * n("HClO")) / (n("ClO"^(-))) = (["OH"^(-)] * ["HClO"]) / (["ClO"^(-)]) * V= K_b * V = 3.4 xx 10^(-7) * 0.102#

Where `n("OH"^(-))`, `n("HClO")`, and `n("ClO"^(-))` the `color(navy)("quantities")` (in number of moles) of each of the respective species present in the solution at its `color(navy)("equilibrium")` position. With reference to the RICE table,

#((x) * (1.2502 xx 10^(-2) + x)) / (1.2998 xx 10^(-2) - x) = 3.448 xx 10^(-7) * 0.102#

Solving this equation for `x` yields `x = 3.6color(grey)(56) xx 10^(-8) color(white)(l) color(purple)(mol)`. Therefore `n("OH"^(-)) = x = 3.6color(grey)(56) xx 10^(-8) color(white)(l) color(purple)(mol)` of `"OH"^(-)` would eventually be present in the solution as it comes to its equilibrium state.

The final solution has a volume of `100.0 + 2.00 = 102.0 color(white)(l) ml` (assuming no volume change due to the mixing.) Thus `["OH"^(-)] = n / V = 3.584 xx 10^(-7) color(white)(l) mol` at equilibrium.

`"pOH" = -log["OH"^(-)] = 6.445`
and therefore
`"pH" = "pKw" - "pOH" = 7.554`.